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(a)To find the turning points of a curve, need to solve dy/dx=0. Using the quotient rule one can differentiate y:y=f(x)/g(x) dy/dx=(f'(x)g(x)-f(x)g'(x))/(g(x))²f(x)=3x, f'(x)=3, g(x)=(9+x^2...

Try differentiating:sin(x)cos(z-x)+cos(x)sin(z-x)with respect to x and for constant z, then simply the derivative you deduce, then integrate throughout the simplified equation. See what happens!

There are two definitions of the sine and cosine functions that anyone who uses contemporary maths, and I do mean anyone, uses silently or otherwise. The first is as follows:'Rotate the point (1,0) in Euc...

Simplify the equation to y=u^3/2 where u = 6x² -5Use the chain ruledy/dx = dy/du x du/dxdy/du = (3/2)u^1/2du/dx = 12x - 0Therefore dy/dx = (3/2)u^1/2 x 12xBut u = ...

3x(x-2)/(2x+1)(x-2)(3x-5) 3x/(2x+1)(3x-5)